7.4Finding the General Term by Comparison

When the differences between terms are not constant, no $T_n = dn + T_0$ fits. Compare the sequence with one you already know, position by position.

Sequences worth knowing:

Sequence	Terms	$T_n$
square numbers	$1, 4, 9, 16, \ldots$	$n^2$
a position times the next position	$2, 6, 12, 20, \ldots$	${n(n + 1)}$
triangular numbers	$1, 3, 6, 10, \ldots$	$\frac{n(n + 1)}{2}$

Take away the known term in each position. If the same amount is left every time, $T_n$ is the known term plus that amount: $3, 6, 11, 18, \ldots$ is $n^2 + 2$ , and $-1, 2, 7, 14, \ldots$ is $n^2 - 2$ .
Divide by the known term. If the answer is the same every time, $T_n$ is that multiple of it: $3, 12, 27, 48, \ldots$ is $3n^2$ .
Divide by the position. If the answers go up by one each time, each term is its position times (its position plus a fixed amount): $4, 10, 18, 28, \ldots$ gives $4, 5, 6, 7$ , so $T_n = n(n + 3)$ .
Square numbers that start later. Write each term as a number squared and compare that number with the position: $16, 25, 36, 49, \ldots$ is $(n + 3)^2$ .

Comparing with the square numbers

The first four terms of a sequence are $\displaystyle 5, 8, 13, 20, \ldots$ Find $\displaystyle T_n$ and hence find $\displaystyle T_{10}$ .

Show solution▾

\displaystyle 5

\displaystyle = 1^2 + 4

\displaystyle 8

\displaystyle = 2^2 + 4

\displaystyle 13

\displaystyle = 3^2 + 4

\displaystyle 20

\displaystyle = 4^2 + 4

Each term is the square of its position, plus 4.

\displaystyle T_n

\displaystyle = n^2 + 4

\displaystyle T_{10}

\displaystyle = 10^2 + 4

\displaystyle = 100 + 4

\displaystyle = 104

The first four terms of a sequence are $\displaystyle 5, 12, 21, 32, \ldots$ Find $\displaystyle T_n$ and hence find $\displaystyle T_9$ .

Show solution▾

Divide each term by its position.

\displaystyle 5 \div 1

\displaystyle = 5

\displaystyle 12 \div 2

\displaystyle = 6

\displaystyle 21 \div 3

\displaystyle = 7

\displaystyle 32 \div 4

\displaystyle = 8

The answers go up by one each time, and each is its position plus 4.

So each term is its position times (its position plus 4).

\displaystyle T_n

\displaystyle = n(n + 4)

\displaystyle T_9

\displaystyle = 9(9 + 4)

\displaystyle = 9 \times 13

\displaystyle = 117

Now you try

The first four terms of a sequence are $3, 6, 11, 18, \ldots$ Find $T_n$ .

Show solution▾

3, 6, 11, 18, \ldots

\displaystyle 3

\displaystyle = 1^2 + 2

\displaystyle 6

\displaystyle = 2^2 + 2

\displaystyle 11

\displaystyle = 3^2 + 2

\displaystyle 18

\displaystyle = 4^2 + 2

\text{Each term is the square of its position } + 2

T_n = n^2 + 2

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